Solving the Mystery of Wavelength Refraction: A Thin 200nm Film of Oil

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SUMMARY

The discussion centers on calculating the wavelength of light reflected from a thin 200nm film of oil (n = 2.0) floating on water (n = 1.33). The correct wavelength, as concluded, is 533 nm, achieved by adjusting the integer m in the interference equation. The initial incorrect calculation of 800 nm stemmed from a misunderstanding of the phase change and the incorrect assumption that m = 1. The final resolution highlights the importance of correctly applying the interference condition for constructive interference.

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  • Understanding of thin film interference
  • Familiarity with the refractive indices of materials
  • Knowledge of the interference equation for constructive interference
  • Basic principles of light wavelength and the visible spectrum
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Homework Statement


http://lon-capa.bd.psu.edu/res/psuerie/gwb6/physics/optics/e1p4.png
A thin 200nm film of oil (n = 2.0) floats on water (n = 1.33). To a normal human, what wavelength will the film appear when viewed from above (nearly perpendicular to the film)?

Homework Equations



for constructive interference
2 * pi * m = ( 2 * pi * n * 2 * d) / lambda + pi (in this case)

The Attempt at a Solution



This was a question I missed on the midterm. I want to know where I went wrong for the final. Here is what I did.

There is a poem our prof taught us; high to low, phase no. Low to high, phase change pi.
Based on the picture, which hopefully shows up, above I can see there is 1 phase change of pi where air meets oil. my equation is then

2 * pi * m = ( 2 * pi * n * 2 * d) / lambda + pi
where m = 1, n = oil = 2, d = 200 nm

I solve for lambda and get the answer of 800 nm.

This is not the correct answer, and I know this right away because light of this wavelength is not in the visible spectrum.

I know the answer is 533 nm but I don't know why. I feel like I am missing something here conceptually.

Please help.So, it looks like the picture didn't make it through to the post it looks something like this the lines separate the different mediums.

air
_________________________

oil
_________________________

water
_________________________
 
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Ok, I figured it out.

First off my math was bad when I came up with the 800 nm. I only multiplied by d rather than 2 d had I done this I would have found a wavelength of 1600 nm. Still not the right answer but it is the right equation.

It then occurred to me that I have a variable that I can play with which is m. I had assumed that m = 1. Bad assumption because there really isn't any reason m can't be any integer I want. If I make m = 2 the problem is solved.
 
JJK1503 said:
There is a poem our prof taught us; high to low, phase no. Low to high, phase change pi.
I like that!

JJK1503 said:
2 * pi * m = ( 2 * pi * n * 2 * d) / lambda + pi
where m = 1, n = oil = 2, d = 200 nm

I solve for lambda and get the answer of 800 nm.
I don't see how you got that answer. Using the same basic approach, I get the given answer.

Ah... I see you've got it now. (Too late! But good for you.)
 

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